The correct answer is $\boxed{\rm{2\hat i} + {\rm{\hat j}}}$.
Two vectors are orthogonal if the dot product of the vectors is zero. The dot product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is defined as follows:
$$\overrightarrow{a} \cdot \overrightarrow{b} = a_x b_x + a_y b_y + a_z b_z$$
where $a_x$, $a_y$, and $a_z$ are the components of $\overrightarrow{a}$, and $b_x$, $b_y$, and $b_z$ are the components of $\overrightarrow{b}$.
Given that $\overrightarrow{\rm{a}} = {\rm{\hat i}} + 2{\rm{\hat j}} + 5{\rm{\hat k}}$ and $\overrightarrow{\rm{b}} = {\rm{\hat i}} + 2{\rm{\hat j}} – {\rm{\hat k}}$, we can calculate the dot product as follows:
$$\overrightarrow{\rm{a}} \cdot \overrightarrow{\rm{b}} = (1)(1) + (2)(2) + (5)(-1) = 0$$
Therefore, $\overrightarrow{\rm{a}}$ and $\overrightarrow{\rm{b}}$ are orthogonal.
The vector $\overrightarrow{\rm{c}}$ is parallel to $\overrightarrow{\rm{a}}$ and $\overrightarrow{\rm{b}}$ if the dot product of $\overrightarrow{\rm{c}}$ and $\overrightarrow{\rm{a}}$ is zero and the dot product of $\overrightarrow{\rm{c}}$ and $\overrightarrow{\rm{b}}$ is zero.
The dot product of $\overrightarrow{\rm{c}}$ and $\overrightarrow{\rm{a}}$ is given by:
$$\overrightarrow{\rm{c}} \cdot \overrightarrow{\rm{a}} = c_x a_x + c_y a_y + c_z a_z$$
The dot product of $\overrightarrow{\rm{c}}$ and $\overrightarrow{\rm{b}}$ is given by:
$$\overrightarrow{\rm{c}} \cdot \overrightarrow{\rm{b}} = c_x b_x + c_y b_y + c_z b_z$$
Since $\overrightarrow{\rm{a}}$ and $\overrightarrow{\rm{b}}$ are orthogonal, we know that $a_x b_x + a_y b_y + a_z b_z = 0$. Therefore, the dot product of $\overrightarrow{\rm{c}}$ and $\overrightarrow{\rm{a}}$ must also be zero in order for $\overrightarrow{\rm{c}}$ to be parallel to $\overrightarrow{\rm{a}}$ and $\overrightarrow{\rm{b}}$.
The only option that satisfies this condition is $\overrightarrow{\rm{c}} = 2{\rm{\hat i}} + {\rm{\hat j}}$. Therefore, the vector $\overrightarrow{\rm{c}}$ is parallel to $\boxed{\rm{2\hat i} + {\rm{\hat j}}}$.